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In Raynaud's "Anneaux locaux henseliens," a proof is given of the following fact: Let $R \to S$ be a finite type morphism of rings, $\mathfrak{q} \in \mathrm{Spec} S$, $\mathfrak{p} $ the inverse image of $\mathfrak{q}$ in $R$, such that $R \to S$ is locally unramified at $\mathfrak{q}$. Then $S$ is isomorphic near $\mathfrak{q}$ to a quotient of a standard etale algebra (i.e. of the form $(R[x]/P)_Q$ where the localization is such that $P'$ is a unit).

In Raynaud's book, Zariski's Main Theorem is first used to reduce to the case of $S$ finite over $R$. The part I don't understand is the reduction to the case of $S$ generated by one element over $R$, where $R$ is local.

Here is the argument, as I understand it. Let the maximal ideal of $R$ be $\mathfrak{m}$ and those of the semi-local ring $S$ be $\mathfrak{q}, \mathfrak{q}_1, \dots, \mathfrak{q}_k$. Choose $x \in S$ generating the residue extension $S/\mathfrak{q}/R/\mathfrak{m}$ (possible by the primitive element theorem, separability following from structure theory of unramified algebras over fields) and in the other maximal ideals $\mathfrak{q}_i$. Consider the algebra $C = R[x] \subset S$ and the ideal $\mathfrak{r} = \mathfrak{q} \cap C$. Raynaud claims on p.53 that $\mathfrak{q}$ is the only maximal ideal of $S$ lying over $\mathfrak{r}$.

I don't understand the claim. Suppose we have an extension of number fields $L/K$ which splits completely at some prime. Let $B/A$ be the ring of integers, and let $\mathfrak{p}$ be the prime of $K$ which splits completely. Then the extension $A_{\mathfrak{p}} \to B_{\mathfrak{p}}$ is unramified and $B_{\mathfrak{p}}$ is a finite $A_{\mathfrak{p}}$-module. However, since the residue extensions are all trivial, we can take $x$ in the above paragraph to be zero. It is, however, not the case that only one prime of $B_{\mathfrak{p}}$ lies over $\mathfrak{p}$, unless the extension is trivial. What am I missing?

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It suffices to impose $x$ to be non-zero in $S/\mathfrak{q}$. –  Qing Liu Jan 2 '11 at 22:17
    
OK, I think I see this argument. If another prime lay above $\mathfrak{r}$, then $x$ would lie in this prime, and so in $\mathfrak{r}$. But it cannot, as it does not lie in $\mathfrak{q}$. –  Akhil Mathew Jan 2 '11 at 23:08
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@Qing Liu: Thanks for pointing that out, by the way! –  Akhil Mathew Jan 2 '11 at 23:12
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The question has been closed at the request of the author. –  Pete L. Clark Jan 2 '11 at 23:44
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Please do not delete it! –  Mariano Suárez-Alvarez Jan 3 '11 at 1:58
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closed as no longer relevant by Akhil Mathew, Andres Caicedo, Yemon Choi, Daniel Moskovich, Pete L. Clark Jan 2 '11 at 23:43

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